Question

QUESTION 1 Let P be a point inside A ABC. Suppose D is the midpoint of AB, E is the midpoint of BC, and F is the midpoint of CA. If PĎ is perpendicular to AB and PE is perpendicular to BC, then PF is perpendicular to CÃ as well. Additionally, make a sketch of a picture labeling all points to illustrate the described setting. In your explanation you can use All Axioms of Incidence, Order, and Congruence and Theorems 1-66.

Answer 1

Given that, In AABC, Point D, E , f are the mid points of Side ABA, BE , CA respectively, And p is a point inside AABC Such that PD & Pe are perpendiculer to AB & B C respectively, We have to show that it is also a perpendicul on à А PD LAB PE IBC if с E of any line Section AD = DB, BE=ēc, BE=ēc, cf = F By perpendiculer Bisector theoram, berpendicular Bisector is equidistance yoom end point of line Section, And PB = PC By transitive property of equality TPA - PC. PA - PB

By Rerpendiculer Pisector otheoram, If any line which is intersecting a line- Section & equidistance from end points of line-Sections then this line will be the Perpendiculer Bisector of line Section Pf LÃ Proved. intersecting ...