IMO 2009 p4, need help in understanding bits of solution

[iron summit]

Kindly help me understand the solution

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Problem: IMO 2009 p4

[iron summit]

so basically we have taken a point K which is the incircle of quad HMCD
can anyone give me some reasoning for why HMK = HDK = 45 degrees

IMO 2009 p4, need help in understanding bits of solution

[green lance]

H is the incentre => CH is the bisector

now use AAS congruence

so HDC and HMC are congruent

so HD and HM are equal

and the angles DHK and MHK are equal

and HK is common, so by SAS congruence, HDK and HMK are congruent

so HDK and HMK are equal

[iron summit]

why 45 degree

so basically I dont understand one thing that is what was the motivation behind making incircle of HDMC

[green lance]

because K is the incentre of ADC

so the line drawn through it is going to bisect the angle

[iron summit]

oh okay

+close

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