Question

In the figure, angle ACB is obtuse and AD is perpendicular to BC produced,D being the foot. Prove that AB² = BC² + CA² +2BC × CD
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Answer 1

Given :-

• ABD a right angled triangle with right angle at D
• ACB a obtuse angled triangle

To Prove :-

• AB² = BC² + CA² +2BC × CD

Proof :-

In ∆ ABD

$\dashrightarrow\sf AB^2= AD^2+BD^2 \ \ \ \ \Big\lgroup By \ Pythagoras\Big\rgroup\\ \\ \\ \bullet \sf Now \ In \ \Delta ADC \\ \\ \\ \dashrightarrow\sf AD^2= AC^2-DC^2------ eq.\ (i)\\ \\ \\ \bullet\sf And \ BD= BC+CD------ eq.\ (ii)\\ \\ \\ \dashrightarrow\sf Substitute\ the \ value of (i) \ and \ (ii)\\ \\ \\\dashrightarrow\sf AB^2= AD^2+BD^2\\ \\ \\ \dashrightarrow\sf AB^2= AC^2-DC^2+(BC+CD)^2\\ \\ \\ \bullet\sf (a+b)^2= a^2+b^2+2ab\\ \\ \\ \dashrightarrow\sf AB^2= AC^2-DC^2+BC^2+CD^2+2BD\times CD\\ \\ \\ \dashrightarrow\sf AB^2= AC^2+BC^2+2BC\times CD\ \ \ \ \ \Big\lgroup \ CD\ will \ be \ cancelled\Big\rgroup\\ \\ \\ \dashrightarrow\sf AB^2=AC^2+BC^2+2BC.CD\ \ \ \ \ ( Hence \ Proved !!)$

Answer 2

The answers is in the attachment.

Please do go through it.

Formulas:

• Pythagoras theorem-In ∆ADC
• AB²=AD²+BD²

• a+b²=a²+b²+2ab.