In Fig. 6.59, ABC is a triangle in which \angle ABC < 90° and AD \perp BC. Prove that AC^2 = AB

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Question 4 In Fig. 6.59, ABC is a triangle in which $\angle ABC < 90°$ and AD $\perp$ BC. Prove that $AC^2 = AB^2 + BC^2 – 2 BC.BD.$

Solution:

By applying Pythagoras Theorem in ∆ADB, we get,

$AB^2 = AD^2 + DB^2$

We can write it as;

$AD^2 = AB^2 − DB^2$ ……………….. (i)

By applying Pythagoras Theorem in ∆ADC, we get,

$AD^2 + DC^2 = AC^2$

From equation (i),

$AB^2 − BD^2 + DC^2 = AC^2$

$AB^2 − BD^2 + (BC − BD)^2 = AC^2$

$AC^2 = AB^2 − BD^2 + BC^2 + BD2 −2BC × BD$

$AC^2 = AB^2 + BC^2 − 2BC × BD$

Hence, proved.

Additional Questions

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Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.	Answer
In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that : (i) Δ APC ~ Δ DPB (ii) AP . PB = CP . DP		Answer
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In Fig. 6.63, D is a point on side BC of Δ ABC such that $\dfrac{BD}{CD} = \dfrac{AB}{AC}$ Prove that AD is the bisector of $\angle BAC$.		Answer
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In Fig. 6.57, D is a point on hypotenuse AC of Δ ABC, such that BD $\perp$ AC, DM $\perp$ BC and DN $\perp$ AB. Prove that : (i) $DM^2 = DN . MC $ (ii) $DN^2 = DM . AN $		Answer
In Fig. 6.58, ABC is a triangle in which $\angle ABC > 90°$ and AD $\perp$ CB produced. Prove that $AC^2 = AB^2 + BC^2 + 2 BC . BD.$		Answer
In Fig. 6.59, ABC is a triangle in which $\angle ABC < 90°$ and AD $\perp$ BC. Prove that $AC^2 = AB^2 + BC^2 – 2 BC.BD.$		Answer

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