Q2 of 49 Page 51

In the adjacent figure, CD and BE are altitudes of an isosceles triangle ABC with AC = AB. Prove that AE = AD.

In Δ ABC BE and CD are the perpendiculars drawn on sides AC and AB respectively


In Δ BDC and Δ BEC


∠BDC = ∠BEC = 900 (Perpendiculars)


∠DBC = ∠ECB(Base angle of the isosceles triangle)


BC = BC(Common side)


So Δ BDC and Δ BEC are congruent to each other by A.A.S. axiom of congruency


DB = EC (Corresponding Part of Congruent Triangle) …(i)


AB = AC (Given) …(ii)


Subtracting (i) from (ii) we get


AB-DB = AC-EC


⇒ AD = AE


Hence Proved


More from this chapter

All 49 →
7

Suppose a triangle is equilateral. Prove that it is equiangular.

 1

In the given figure, If AB || DC and P is the midpoint of BD, prove that P is also the midpoint of AC.

 3

In the figure, AP and BQ are perpendiculars to the line segment AB and AP = BQ. Prove that O is the midpoint of line segment AB as well as PQ.

 4

Suppose ABC is an isosceles triangle with AB = AC; BD and CE are bisectors of ∠B and ∠C. Prove that BD = CE.