Calculate Angle CDB in an Isosceles Right Triangle with Median

Q: Why is the angle 90° and not 45°?

The median BD goes to the midpoint of the hypotenuse AC, not to split an angle! In any right triangle, the median to the hypotenuse creates perpendicular angles of 90° at point D.

Q: What's the difference between a median and an angle bisector?

A median connects a vertex to the midpoint of the opposite side. An angle bisector splits an angle in half. They're completely different! Here, BD is a median, so it creates 90° angles.

Q: Does this work for all right triangles or just isosceles ones?

This property works for all right triangles! The median to the hypotenuse always creates 90° angles at the midpoint, whether the triangle is isosceles or not.

Q: How can I remember this property?

Think: "Median to hypotenuse makes right angles". The median creates a perpendicular from the right angle vertex to the hypotenuse, always giving you 90°!

Q: What if the triangle wasn't a right triangle?

If triangle ABC wasn't a right triangle, then $ \angle CDB $ would not be 90°. This special property only works when the original triangle has a right angle.

Median Properties with Right Triangle Angles

ABC is an isosceles right triangle.

BD is the median.
How large is angle $∢\text{CDB}$ ?

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:08 Let's find angle C D B.

00:12 We have an isosceles triangle in the diagram.

00:16 In isosceles triangles, the base angles are equal.

00:22 Angles in triangle A B C add up to 180 degrees.

00:27 Since it's an isosceles triangle, the base angles are the same.

00:37 Let's focus on finding angle A.

00:42 This gives us the size for angles A and C.

00:51 Line B D is a median of the triangle.

00:55 In isosceles triangles, the median is also a height.

01:06 And that's how we solve this problem.

Step-by-step written solution

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Understand the problem

ABC is an isosceles right triangle.

BD is the median.
How large is angle $∢\text{CDB}$ ?

Step-by-step solution

To solve this problem, we must understand the properties of an isosceles right triangle. In $\triangle ABC$ , since it is an isosceles right triangle: $\angle ACB = \angle CAB = 45^\circ$ and $\angle ABC = 90^\circ$ .

The median BD will divide the triangle into two smaller triangles, ABD and CBD. In $\triangle ABC$ , since BD is a median from vertex B to side AC, and because AC is the hypotenuse of this right triangle, BD is equal in length to segments AD and DC. By the definition of the median and symmetry of the isosceles triangle, angles $\angle ADB$ and $\angle CDB$ are both right angles. Therefore, $\angle CDB$ is $90^\circ$ .

Consequently, the measure of angle $\angle CDB$ is clearly $90^\circ$ .

Therefore, the solution to the problem is $\angle CDB = 90^\circ$ .

Final Answer

Key Points to Remember

Essential concepts to master this topic

Property: Median to hypotenuse equals half the hypotenuse length
Technique: In right triangle, median creates two right angles at midpoint
Check: Verify $\angle CDB = 90°$ using isosceles triangle symmetry ✓

Common Mistakes

Avoid these frequent errors

Assuming the median creates 45° angles
Don't think the median splits the vertex angle in half = 45° angles! This confuses angle bisectors with medians. Always remember that a median to the hypotenuse of a right triangle creates perpendicular 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 is the angle 90° and not 45°?

The median BD goes to the midpoint of the hypotenuse AC, not to split an angle! In any right triangle, the median to the hypotenuse creates perpendicular angles of 90° at point D.

What's the difference between a median and an angle bisector?

A median connects a vertex to the midpoint of the opposite side. An angle bisector splits an angle in half. They're completely different! Here, BD is a median, so it creates 90° angles.

Does this work for all right triangles or just isosceles ones?

This property works for all right triangles! The median to the hypotenuse always creates 90° angles at the midpoint, whether the triangle is isosceles or not.

How can I remember this property?

Think: "Median to hypotenuse makes right angles". The median creates a perpendicular from the right angle vertex to the hypotenuse, always giving you 90°!

What if the triangle wasn't a right triangle?

If triangle ABC wasn't a right triangle, then $\angle CDB$ would not be 90°. This special property only works when the original triangle has a right angle.

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