## Saturday, July 22, 2023

### Geometry Problem 1548: Exploring Segment DE Length in Triangle ABC with a 45-Degree Angle and Intersecting Altitudes

Geometry Problem 1548. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

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1. Let BF = p so that BE = EF = p/V2
Let CF = q so that CD = q/V2

Triangles AGE & BCD are similar, hence
36/AE = BD/a so 36/(b/V2) = c/V2/(a) from which
bc = 72a..............(1)

Triangles EFH & BCD are similar. hence
6V2/p = q/V2/a so
pq = 12a.......(2)

Triangles AED & ABC are similar, hence
DE = a/V2.......(3)

Now apply Ptolemy's Theorem to cyclic quadrilateral BCDE
a.DE + (p/V2)(q/V2) = (b/V2).(c/V2) and so
a^2/V2 + pq/2 = bc/2 from which
a^2/V2 + 6a = 36a where we have substituted from (1) & (2)

Therefore from (3) DE = a/V2 = 30

Sumith Peiris
Moratuwa
Sri Lanka

1. Much simpler solution

Draw EPQ // BD, P on AG & Q on AC

BCDE is concyclic hence < AEG = C and < EAP = 90 - C = < CBD = < DEC
Consider Triangles AEP & CED

*AE = CE
*< EAP = < DEC = 90 - C
*< AEP = < < DCE = 45

So Triangles AEP & CED are congruent ASA

Hence AP = DE ......(1) and
EP = DC = FD since Tr. CDF is right isosceles

So Triangles EPG & DFH are congruent ASA
Hence PG = FH = 6 .....(2)

From (1) & (2), DE = AP = AG - PG = 36 - 6
DE = 30

Sumith Peiris
Moratuwa
Sri Lanka

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