Calculate Angle ADB in an Isosceles Right Triangle with Median

Q: Why isn't angle ADB equal to 45° like the other angles?

Great question! While angles A and C are 45°, angle ADB is different because D is on the hypotenuse. The median to the hypotenuse creates a special perpendicular relationship, making ∠ADB = 90°.

Q: What makes BD a median and not just any line?

A median connects a vertex to the midpoint of the opposite side. Since BD goes from vertex B to point D (midpoint of AC), it's specifically the median from B to the hypotenuse.

Q: How do I know that triangles ABD and BDC are congruent?

They're congruent because: AD = DC (D is midpoint), BD = BD (shared side), and both are right triangles with equal legs from the isosceles property. This gives us SSS congruence!

Q: Is this 90° angle property true for all right triangles?

Not quite! This special property only works for isosceles right triangles. In other right triangles, the median to the hypotenuse doesn't necessarily create a 90° angle.

Q: Can I solve this using coordinate geometry instead?

Absolutely! Place B at origin, A at (0,a), and C at (a,0). Then D is at $ (\frac{a}{2}, \frac{a}{2}) $, and you can calculate that the angle ADB is indeed 90° using dot products.

Angle Properties with Median Construction

ABC is an isosceles right triangle.

BD is the median.

Calculate the size of angle $∢ADB$ .

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Step-by-step video solution

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00:08 Let's find the angle A D B.

00:11 We have an isosceles triangle based on the information given.

00:15 B D is a median, according to the data provided.

00:19 In an isosceles triangle, the median is also the height.

00:25 Here's how we solve it!

Step-by-step written solution

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Understand the problem

ABC is an isosceles right triangle.

BD is the median.

Calculate the size of angle $∢ADB$ .

Step-by-step solution

The solution to find $\angle ADB$ can be determined by understanding the properties of the isosceles right triangle and the median.

Given triangle $\triangle ABC$ is an isosceles right triangle with $\angle ABC = 90^\circ$ and $\angle BAC = \angle ACB = 45^\circ$ , BD, being the median from B to the hypotenuse AC, causes A and C to be equally distant from D. Consequently, triangle ABD is congruent to triangle BDC.

In this configuration, triangle ABD forms half of the isosceles right triangle. Due to symmetry and the properties of medians in such triangles, we have two equal angles at ADB and BDC.

Since $\triangle ABD$ and $\triangle BDC$ are congruent under isosceles conditions, we find that angle $\angle ADB$ , like $\angle ABC$ , must also be $\angle ADB = 90^\circ$ .

Therefore, the solution to $\angle ADB$ is $\boxed{90}$ degrees.

Final Answer

Key Points to Remember

Essential concepts to master this topic

Isosceles Rule: Right triangle at B creates 45° angles at A and C
Median Property: BD to hypotenuse makes D the midpoint of AC
Check: Triangle ABD and BDC are congruent, confirming ∠ADB = 90° ✓

Common Mistakes

Avoid these frequent errors

Assuming angle ADB equals 45° like base angles
Don't think ∠ADB = 45° just because ∠BAC and ∠ACB are 45° = wrong triangle properties! The median creates a special perpendicular relationship. Always recognize that medians to the hypotenuse in right triangles create 90° angles at the midpoint.

Practice Quiz

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Is DE side in one of the triangles?

FAQ

Everything you need to know about this question

Why isn't angle ADB equal to 45° like the other angles?

Great question! While angles A and C are 45°, angle ADB is different because D is on the hypotenuse. The median to the hypotenuse creates a special perpendicular relationship, making ∠ADB = 90°.

What makes BD a median and not just any line?

A median connects a vertex to the midpoint of the opposite side. Since BD goes from vertex B to point D (midpoint of AC), it's specifically the median from B to the hypotenuse.

How do I know that triangles ABD and BDC are congruent?

They're congruent because: AD = DC (D is midpoint), BD = BD (shared side), and both are right triangles with equal legs from the isosceles property. This gives us SSS congruence!

Is this 90° angle property true for all right triangles?

Not quite! This special property only works for isosceles right triangles. In other right triangles, the median to the hypotenuse doesn't necessarily create a 90° angle.

Can I solve this using coordinate geometry instead?

Absolutely! Place B at origin, A at (0,a), and C at (a,0). Then D is at $(\frac{a}{2}, \frac{a}{2})$ , and you can calculate that the angle ADB is indeed 90° using dot products.

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