ABC is a triangle in which AB = AC and D is any point in BC. Prove that : (AB)2 – (AD)2 = BD . CD.

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Draw the perpendicular from A to BC. Let it intersect BC at E. Then in triangle AEB we have: AB^2 = AE^2 + BE^2 In triangle AED we have: AD^2 = AE^2 + BD^2 Subtracting gives: AB^2 - AD^2 = BE^2 - BD^2 AB^2 - AD^2 = (BE + BD)(BE - BD) Now, the perpendicular to the unequal side in an isosceles...
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Draw the perpendicular from A to BC. Let it intersect BC at E. Then in triangle AEB we have: AB^2 = AE^2 + BE^2 In triangle AED we have: AD^2 = AE^2 + BD^2 Subtracting gives: AB^2 - AD^2 = BE^2 - BD^2 AB^2 - AD^2 = (BE + BD)(BE - BD) Now, the perpendicular to the unequal side in an isosceles triangle is also the bisector of the unequal side. This can be proved by considering triangles AEB and AEC AB = AC angle AEB = angle AEC AE = AE By RHS criterion, the two triangles are congruent. So BE = CE Depending on where you draw point E, you can get one of two things: BE + ED = BD and BE - ED = CE - ED = CD (or) BE + ED = CE + ED = CD and BE - ED = BD In either case you get: (BE+ED)*(BE-ED) = BD*CD. Thus: AB^2 - AD^2 = BD*CD read less
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