Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Erina-NJ.

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## Friday, November 30, 2018

### Geometry Problem 1403: Quadrilateral, 90 Degree Angle, Perpendicular, Congruence

Labels:
90,
angle,
congruence,
geometry problem,
perpendicular,
quadrilateral

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Let AN and DM intersect at point Q, then we have AQ.AN=AD^2.

ReplyDeleteLet perpendicular drawn from B to AM meets AM at P, then we have AP.AM=AB^2=AD^2

We get AQ.AN=AP.AM, hence Quad.MPQN is cyclic,Angle MQN= Angle MPN=90 deg.

Since Angle BPM = Angle MPN = 90 Deg, B,P and N are collinear points and BN is perpendicular to AM.

Let AN, DM cut at X & BN, AM cut at Y.

ReplyDeleteAX.AN = AD^2 = AB^2 so < BNX = < ABX = < AMX since ABMX is concyclic

Hence MNXY is concyclic and so

< MYN = < MXN = 90

Sumith Peiris

Moratuwa

Sri Lanka