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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
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B
False
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Solution

The correct option is A True
Due 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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