Abc is an isosceles triangle in …
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Sia ? 4 years, 10 months ago
Construction : Draw {tex}A L \perp B C{/tex}
or, AL is median of BC (Isosceles triangle)
or, BL = LC = 6 cm.
In right {tex}\triangle A L B,{/tex} by Pythagoras theorem, AL = 8 cm.
In {tex}\triangle B P Q \text { and } \triangle B L A{/tex}
{tex}\angle B = \angle C{/tex} (Isosceles triangle)
{tex}\angle B P Q = \angle B L A = 90 ^ { \circ }{/tex}
{tex}\triangle B P Q \sim \triangle B L A{/tex} (AA similarity)
or, {tex}\frac { B P } { P Q } = \frac { B L } { A L }{/tex}
or, {tex}\frac { 6 - x } { y } = \frac { 6 } { 8 } \text { or, } x = 6 - \frac { 3 y } { 4 }{/tex}
Hence proved.
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