Question

Q. The altitude AD of triangle ABC is produced to cut it's circumcircle in 'K'. Prove that HD = DK, where 'H' is the orthocentre.

Answer 1

♦ Find the right angles : )

( 1 ) ∠HBC = 90° - ∠ACB 

=> ∠BHD = 90° - ∠HBC = ∠ACB

( 2 ) ∠BKH = ∠BCA   [ Angles in the same segment ]

=> ∠BHK = ∠BHD = ∠BKH = ∠ACB

=> ΔBHK is isosceles

→ ∠D = 90°
=> BD is ⊥ on KH

=> HD = DK [ Proved ]

Answer 2

Hey !!

Here is your answer.. ⬇⬇

➡ Given :- AD and BE are the altitudes of ∆ABC.

➡ To Prove :- HD = DK

➡ Proff :-

angle KBC = angle KAC --( Angle lies in a same segment of circle ) -- ( 1 )

↪ In ∆ADC, AD is perpendicular to BC. ( ADC = 90° )

angle ADC + angle ACD + angle CAD = 180

90 + angle ACD + angle CAD = 180

Angle CAD = 180 - 90 - angle ACD

angle CAD = 90 - angle ACD

Angle CAK = 90 - angle ACD -- ( CAD = CAK ) --(2 )

angle KBC = 90 - angle ACD -- ( from eq.1 ) ( 3 )

↪ In ∆CBE, BE perpendicular AC ( BEC = 90° )

Angle CBE + angle BCE + angle BEC = 180

90 + angle CBE + angle BCE = 180

Angle CBE = 90 - angle BCE --( 4 )

↪ From eq. ( 2 ), ( 3 ), and ( 4 ) ..

KAC = KBC = CBE - ( 5 )

↪ In ∆HBD and ∆KDB..

Angle BDH = angle BDK = 90

angle KBD = angle HBD -- ( from eq.5 )

BD = BD -- ( Common )

Therefore, ∆HBD ~ ∆KDB

Hence, HD = DK -- ( By CPCT )

HOPE IT HELPS

THANKS ^-^