In an isosceles triangle ABC with AB = AC

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Question:

In an isosceles triangle ABC with AB = AC and BD ⊥ AC. Prove that BD2 − CD2 = 2CD.AD.

Solution:

Given: ∆ABC is an isosceles triangle with AB = AC and BD ⊥ AC.

To prove: BD2 − CD2 = 2CD.AD

Proof:  In right ∆ADB,

$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$ (Pythagoras Theorem)

$\Rightarrow \mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2} \quad(\mathrm{AB}=\mathrm{AC})$

$\Rightarrow(\mathrm{AD}+\mathrm{CD})^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$

$\Rightarrow \mathrm{AD}^{2}+\mathrm{CD}^{2}+2 \mathrm{AD} \cdot \mathrm{CD}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$

$\Rightarrow \mathrm{BD}^{2}-\mathrm{CD}^{2}=2 \mathrm{AD} \cdot \mathrm{CD}$      (Hence Proved)

 

 

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