Question

PLEASE SOLVE FAST...​

Answer 1

$\displaystyle\sf{Given\:,$

$\displaystyle\sf{\triangle ABC \:and\:altitude\:AD\:and\:CE\:of\:triangle\:intersects\:each\:other\:at\:point\:F}$

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$\displaystyle\sf{(i)\:\triangle AEF \sim \triangle CDF$

$\displaystyle\sf{In\:\triangle AEF \:and\: \triangle CDF\:,$

• $\displaystyle\sf{<AEF=\:<CDF\:(since\:AD\:and\:CE\:are\:altitudes)$

• $\displaystyle\sf{<AFE=\:<CFD\:(Vertically\:opposite\:angles)$

$\displaystyle\sf{Hence\:\triangle AEF \sim \triangle CDF\:(AA\:similarity)$

$\displaystyle\sf{\underline{Hence\:Proved\:!\:!\:!}}$

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$\displaystyle\sf{(ii)\:\triangle ABD \sim \triangle CBE$

$\displaystyle\sf{In\:\triangle ABD \:and\: \triangle CBE\:,$

• $\displaystyle\sf{<ABD=\:<CBE\:(Common\:angle)$

• $\displaystyle\sf{<ADB=\:<CEB\:(Since\:AD\:and\:CE\:are\:altitudes\:,\:both\:90^o)$

$\displaystyle\sf{Hence\:\triangle ABD \sim \triangle CBE\:(AA\:similarity)$

$\displaystyle\sf{\underline{Hence\:Proved\:!\:!\:!}}$

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$\displaystyle\sf{(iii)\:\triangle AEF \sim \triangle ADB$

$\displaystyle\sf{In\:\triangle AEF \:and\: \triangle ADB\:,$

• $\displaystyle\sf{<FAE=\:<DAB\:(Common\:angle)$

• $\displaystyle\sf{<AEF=\:<ADB\:(Since\:AD\:and\:CE\:are\:altitudes\:,\:both\:90^o)$

$\displaystyle\sf{Hence\:\triangle AEF \sim \triangle ADB\:(AA\:similarity)$

$\displaystyle\sf{\underline{Hence\:Proved\:!\:!\:!}}$

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$\displaystyle\sf{(iv)\:\triangle FDC \sim \triangle BEC$

$\displaystyle\sf{In\:\triangle FDC \:and\: \triangle BEC\:,$

• $\displaystyle\sf{<BCE=\:<FCD\:(Common\:angle)$

• $\displaystyle\sf{<CEB=\:<CDF\:(Since\:AD\:and\:CE\:are\:altitudes\:,\:both\:90^o)$

$\displaystyle\sf{Hence\:\triangle FDC \sim \triangle BEC\:(AA\:similarity)$

$\displaystyle\sf{\underline{Hence\:Proved\:!\:!\:!}}$