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Area of a Circle

Let ABC be ...

Question

Let $A B C$ be an acute triangle whose orthocenter is at $H$ . If altitude from A is produced to meet the circumcircle of triangle ABC at D, then find $H D$

4 R cos B cos C

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4 cos B cos C

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4 R cos B

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R cos B cos C

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Solution

The correct option is D $4 R cos B cos C$
Consider, $△ B H N$ and $△ B D N$
$∠ C = ∠ C$
$B N = B N$
$∠ B N H = ∠ B N D$ [Right-angled angle]

$△ B H N$ and $△ B D N$ are congurent

hence , $H N = N D = 2 R cos B cos C$

$H D = 2 H N$

$H D = 4 R cos B cos C$

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Let ABC be an acute triangle whose orthocenter is at H. If altitude from A is produced to meet the circumcircle of triangle ABC at D, then find HD

The correct option is D 4RcosBcosCConsider, △BHN and △BDN ∠C=∠C BN=BN∠BNH=∠BND [Right-angled angle]△BHN and △BDN are congurenthence , HN=ND=2RcosBcosCHD=2HNHD=4RcosBcosC

Let $A B C$ be an acute triangle whose orthocenter is at $H$ . If altitude from A is produced to meet the circumcircle of triangle ABC at D, then find $H D$

The correct option is D $4 R cos B cos C$
Consider, $△ B H N$ and $△ B D N$
$∠ C = ∠ C$
$B N = B N$
$∠ B N H = ∠ B N D$ [Right-angled angle]

$△ B H N$ and $△ B D N$ are congurent

hence , $H N = N D = 2 R cos B cos C$

$H D = 2 H N$

$H D = 4 R cos B cos C$