In triangle ABC, angle A = angle C, and AB = AC. D is a point on side AC such that BD is perpendicular to AC. Prove that triangle ABD is isosceles.
Question
In triangle ABC, angle A = angle C, and AB = AC.
D is a point on side AC such that BD is perpendicular to AC.
Prove that triangle ABD is isosceles.
Solution
To prove that triangle ABD is isosceles, we need to show that AD = BD.
Given that triangle ABC is isosceles (since AB = AC and angle A = angle C), we know that the base angles of an isosceles triangle are equal. Therefore, angle ABC = angle ACB.
Since BD is perpendicular to AC, angle ABD = angle ABC = 90 degrees.
Similarly, angle ADB = angle ACB = 90 degrees.
Now, we have two triangles ABD and ADB with:
- A common side AD.
- Angle ABD = angle ADB = 90 degrees.
- Angle BAD = angle BDA (since they are both equal to angle BAC).
By the Angle-Angle-Side (AAS) criterion, we can say that triangle ABD is congruent to triangle ADB.
Since corresponding parts of congruent triangles are equal, we have AD = BD.
Therefore, triangle ABD is isosceles.
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