$△$ ABC is a right angled triangle, right angled at B. BD is a perpendicular as shown. Prove that:

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

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Solution

Since ΔADB and ΔBDC are right-angled triangles, using Pythagoras theorem, we can write:

In triangle ABD,
${A B}^{2}$ = ${A D}^{2}$ + ${B D}^{2}$

In triangle BDC
${B C}^{2}$ = ${B D}^{2}$ + ${D C}^{2}$
=> ${B D}^{2}$ = ${B C}^{2}$ - ${D C}^{2}$

Replacing this value of ${B D}^{2}$ in the first equation, we get:

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

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Similar questions

$△$ 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 perpendicular as shown. If AD: DC = 1:2, then $\frac{{A B}^{2}}{{B C}^{2}}$ is:

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{B C}^{2} \times A D = {A C}^{2} \times B D

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