In -ABC- BD -AC- D -AC and -B is a right angle- If AC -5CD then show that BD -2CD-

From #160; the triangle BDC. BC ^ 2 = BD ^ 2 + x ^ 2 ...(i) andFrom #160; the triangle ABD AB ^ 2 = BD ^ 2 + 16x ^ 2 ...(ii) From #160; the triang ...

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Standard VII

Mathematics

Classification of Triangles Based on Angles

In ΔABC, BD...

Question

In $Δ$ ABC, BD $⊥$ AC, D $\in$ AC and $∠$ B is a right angle. If AC $= 5$ CD then show that BD $= 2$ CD.

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Solution

From the triangle BDC.
${B C}^{2} = {B D}^{2} + x^{2} . . . (i)$
and
From the triangle ABD
${A B}^{2} = {B D}^{2} + {16 x}^{2} . . . (i i)$
From the triangle ABC
${A B}^{2} + {B C}^{2} = {A C}^{2} . . . (i i i)$
Thus substituting in (iii) from (i) and (ii),
${B D}^{2} + x^{2} + {B D}^{2} + 16 x^{2} = (25 x^{2}) 2 {B D}^{2} = 8 x^{2} {B D}^{2} = 4 x^{2} B D = 2 x$
hence
$B D = 2 C D$

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In ΔABC, BD⊥AC, D∈AC and ∠B is a right angle. If AC=5CD then show that BD=2CD.

From the triangle BDC.BC2=BD2+x2...(i)andFrom the triangle ABDAB2=BD2+16x2...(ii)From the triangle ABCAB2+BC2=AC2...(iii)Thus substituting in (iii) from (i) and (ii),BD2+x2+BD2+16x2=(25x2)2BD2=8x2BD2=4x2BD=2xhenceBD=2CD

In $Δ$ ABC, BD $⊥$ AC, D $\in$ AC and $∠$ B is a right angle. If AC $= 5$ CD then show that BD $= 2$ CD.