If any point D is taken on the base BC of an isosceles triangle ABC and DEF is drawn perpendicular to the base BC and meets AB and AC or produced in E and F, show that (DE + DF) is a constant quantity and equals twice the perpendicular from A to BC. CONSTRUCTION: Draw AP and AQ Is BC and DEF respectively. From B draw BF' | BC to meet CA produced in F' A E B D

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.3: Lines
Problem 22E
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If any point D is taken on the base BC of an isosceles triangle ABC and
DEF is drawn perpendicular to the base BC and meets AB and AC or produced
in E and F, show that (DE + DF) is a constant quantity and equals twice
the perpendicular from A to BC.
CONSTRUCTION: Draw AP and AQ Is BC and DEF respectively.
From B draw BF' | BC to meet CA produced in F'
A
E
B
D
Transcribed Image Text:If any point D is taken on the base BC of an isosceles triangle ABC and DEF is drawn perpendicular to the base BC and meets AB and AC or produced in E and F, show that (DE + DF) is a constant quantity and equals twice the perpendicular from A to BC. CONSTRUCTION: Draw AP and AQ Is BC and DEF respectively. From B draw BF' | BC to meet CA produced in F' A E B D
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