## Wednesday, October 5, 2016

### Geometry Problem 1272 Isosceles Triangle, Median, Midpoint, Perpendicular, 90 Degrees, Angles, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1272.

1. https://goo.gl/photos/bqyrjE7bstYM7kCDA

Draw rectangle AMNF ( see sketch)
Note that E is the midpoint of AB and FN.
∠NAM=∠MCN( ANC is isoceles)
And ∠NAM=∠FMA( AMNF os a rectangle)
So ∠FMA=∠MCN => FM//NC => ∠FMC=∠NDM= 90 degrees
In trapezoid FMDN , since MD ⊥ FM and E is the midpoint of NF so EM=ED

2. Problem 1272
Is <AMD=<CNB (they have their sides perpendicular).But AM/MD=MC/MD=NC/NM=NC/NB (AM=MC, NM=NB, triangle MCD is similar with triangle NCM).Therefore triangle AMD is similar with triangle CNB .So <MAD=<NCB.Therefore <DAB=<MCN (<BAC=<BCA ).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

3. Area of Tr. MNC = ½ MD.NC = ½ MN.MC…(1)

But AN = NC, NB =MN and AM = MC

So from (1), MD.AN = NB.AM
Hence in Tr.s ANB and AMD,
AN/NB = AM/MD.

Further the included angles
< ANB = < AMD
(since < ANM = < CNM = < CMD)

So it follows that the Tr.s ANB and AMD are similar and therefore

Hence AMDB is concyclic and so
< BAD = < BMD = < CAN.

Sumith Peiris
Moratuwa
Sri Lanka

4. Peter's comment - "In trapezoid FMDN , since MD ⊥ FM and E is the midpoint of NF so EM=ED"

This is so since the perpendicular bisector of MD passes thro' E

5. Sumith