Question

the prependicular from A on the side BC of a angle ABC intersects BC at D such that DB=3CD. Prove that 2AB^2=2AC^2+2BC^2​

Answer 1

Let BD = 3y , CD = y, and thus BC = 4y.

Applying Pythagoras theorem,

In ∆ABD , AB² = AD²+BD²

In ∆ACD, AC² = AD²+CD²,

from equations above equate for AD, we get,

AB²–BD² = AC²–CD²

AB² = AC² + 9y²–y²

= AC² + 8y²

multiply by 2,

2AB² = 2AC²+16y²

which is,

2AB² = 2AC² + BC²

Hope it helps you my dear friend

Answer 2

$<b> <u> Heya ! </u> </b>$

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★$<u> Triangles </u>$ ★

$<b>Ques.→ The perpendicular from A on the side BC of a triangle ABC intersects BC at D such that DB=3CD. Prove that 2AB²=2AC²+2BC² </b>$

Solution → $<i> Refer the Attachment for the figure .</i>$

$<b>Given </b>$: In ∆ABC , AD _|_ BC

DB = 3 CD .

$<b> <u>To Prove</u> </b>$ : 2AB² = 2AC² + BC²

Proof :-

→ In Right angle triangles , ABD and ADC we have :

=> AB² = AD² + BD² [1.]

=> AC² = AD² + DC² [2]

{ By Pythagoras Theorem }

★ Subtract equation [2] - [1]

=> AB² - AC²= BD² - DC²

[ °•° BD = 3CD]

=> 9CD² = 8 ( BC/4)²

=> AB² - AC² = BC²/2

•°• 2AB² - 2AC² = BC²



Thanks !!
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