Q.1Triangle ABC is isoceles with AB=AC. D, E and F are midpoints of BC, AC and AB respectively. Show that line segments AD is perpendicular to the line segment EF and is bisected by it?
Q.2 If a diagonal of a parallelogram bisects one of the angles of the parallelogram, it also bisects the second angle and then the two diagonals are perpendicular to each other?
given: ABC is a triangle. and D, E and F are the mid-points of BC , AC and AB. and AB=AC
TPT: AD?EF and AG bisects EF.
proof: construction: join EF.
in the triangle ADB and ADC,
BD=DC (D is the mid-point)
and AD is common.
therefore by SSS congruency triangle ADB is congruent to triangle ADC.
therefore ?BAD=?CAD therefore?FAG=?EAG
now in the triangle FAG and triangle EAG,
AG is common.
and AF=AE [since E and F are the mid-points of AB and AC and AB=AC]
therefore by SAS congruency triangle FAG and triangle EAG are congruent.
hence by congruent property : FG=GE and ?AGF=?AGE=180/2=90 deg
hence AD bisects EF and perpendicular to EF.
(2) given: PQRS is a parallelogram. and diagonal PR divides the ?P into two equal parts.
TPT: PR bisects the ?R and diagonals PR and QS are perpendicular.
proof: let PR and QS intersect at point O.
since PS and QR are parallel lines. and PR is transversal line.
similarly for the set of parallel lines PQ and SR,
?QPR=?PRS=x, therefore ?PRQ=?PRS=x
hence PR bisects the ?R also.
in triangle PSR, ?SPR=?PRS=x
therefore PS=SR [sides opposite to equal angles are equal in length]
therefore PQRS is a rhombus.
and the diagonals of a rhombus are perpendicular to each other.