Question

# $$\triangle ABC$$ is an isosceles triangle in which $$AB = AC$$ and $$D$$ is a point on $$BC$$.  then that relation is $$AB^{2} - AD^{2} = BD \times CD$$. ?

A
True
B
False

Solution

## The correct option is A TrueDue to being an isosceles triangle the sides $$AB = AC$$.$$AE$$ is perpendicular to $$BC$$ as $$AE$$ is acting as a median for the triangle $$ABC$$ which is isosceles in nature.  Therefore, we can say that  $$DE + CE = DE + BE$$. Hence,$$AD^2=AE^2+DE^2$$\$  therefore to find $$AE^2=AD^2-DE^2$$In triangle, the value of $$AC$$ is equal to $$\sqrt{AE^2+EC^2}$$Therefore, using the different approach of $$AE$$ i.e $$AE^2=(AD^2-DE^2 and AC^2-EC^2)$$We can say that $$AD^2-DE^2=AC^2-EC^2$$Hence, solving the value of $$AC^2-AD^2=EC^2-DE^2$$we get $$DE^2-EC^2$$ as $$(DE+EC)(DE-EC)$$Now$$(DE+EC)(DE-EC)$$  can also be written as $$CD\times BD$$ according to the $$DE + CE = DE + BE$$.  Therefore,$$AC^2-AD^2=CD\times BD$$ Hence, Proved.Maths

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