ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.
Solution:
Given, ABC is an isosceles triangle with AC = BC
AD and BE are two altitudes to the sides BC and AC.
We have to prove that AE = BD
Considering triangle ABC,
Given, AC = BC
We know that the angles opposite to the equal sides are equal.
∠ABC = ∠CAB
Also, ∠ABD = ∠EAB ------------------- (1)
Considering triangles AEB and BDA,
AD is the altitude of BC
So, AD ⊥ BC
This implies ∠D = 90°
BE is the altitude of AC
So, BE ⊥ AC
This implies ∠E = 90°
Now, ∠AEB = ∠ADB = 90°
From (1), ∠ABD = ∠EAB
Common side = AB
By AAS (Angle-angle-side) criterion, the triangles AEB and BDA are congruent
The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are congruent, then their corresponding sides and angles are also congruent or equal in measurements.
By CPCT,
AE = BD
Therefore, it is proved that AE = BD.
✦ Try This: In △ABC, AD is perpendicular bisector of BC. Show that △ABC is an isosceles triangle in which AB=AC
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 9
ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD
Summary:
ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. It is proven that AE = BD by AAS criterion
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