$△$ $A B C$ is a right angled triangle, right angled at $B$ . $B D$ is a perpendicular as shown. Prove that:

${A B}^{2}$ = ${A D}^{2}$ + ${B C}^{2}$ - ${C D}^{2}$

[3 Marks]

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Solution

Since Δ $A D B$ and Δ $B D C$ are right-angled triangles, using Pythagoras theorem, we can write:

In triangle $A B D$ ,
${A B}^{2}$ = ${A D}^{2}$ + ${B D}^{2}$ ......(i)

[1 Mark]

In triangle $B D C$ ,
${B C}^{2}$ = ${B D}^{2}$ + ${D C}^{2}$
$\Rightarrow {B D}^{2}$ = ${B C}^{2}$ - ${D C}^{2}$ ......(ii)

[1 Mark]

Replacing this value of ${B D}^{2}$ in (i) equation, we get:

${A B}^{2}$ = ${A D}^{2}$ + ${B C}^{2}$ - ${C D}^{2}$
Hence Proved.

[1 Mark]

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Similar questions

$Δ A B D$ is a right triangle right-angled at A and AC \perp BD. Show that
(i) $A B^{2} = B C . B D$
(ii) $A C^{2} = B C . D C$
(iii) $A D^{2} = B D . C D$
(iv) $\frac{A B^{2}}{A C^{2}} = \frac{B D}{D C}$

$△$ ABC is a right angled triangle, right angled at B. BD is a perpendicular as shown. Which of the following is true?

$△$ ABC is a triangle, right angled at B. BD is a perpendicular to AC as shown. Which of the following is true?

△

ABC is a right angled triangle, right angled at B. BD is a perpendicular to AC as shown in the figure. If AD:DC = 1:2, then

\frac{{A B}^{2}}{{B C}^{2}}

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△ABC is a right angled triangle, right angled at B. BD is a perpendicular as shown. Prove that: AB2 = AD2 + BC2 - CD2 [3 Marks]

$△$ $A B C$ is a right angled triangle, right angled at $B$ . $B D$ is a perpendicular as shown. Prove that:

${A B}^{2}$ = ${A D}^{2}$ + ${B C}^{2}$ - ${C D}^{2}$

[3 Marks]