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)
AB=BC
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,
?FAG=?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.
hence proved
(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.
let ?QPO=?OPS=x
since PS and QR are parallel lines. and PR is transversal line.
?PRQ=?RPS=x
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.