Advertisements
Advertisements
Question
AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.
Advertisements
Solution
In ΔAOD and ΔBOC,
∠ AOD = ∠ BOC ....(vertically opposite angles)
∠ DAO = ∠ CBO ....(each 90°)
AD = BC ....(given)
∴ ΔAOD ≅ ΔBOC ...(by AAS congruence criterion)
⇒ AO = BO ...(c.p.c.t.)
⇒ O is the mid-point of AB.
Hence, CD bisects AB.
APPEARS IN
RELATED QUESTIONS
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that
- ΔDAP ≅ ΔEBP
- AD = BE
Which congruence criterion do you use in the following?
Given: ∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ΔLMN ≅ ΔGFH
You want to show that ΔART ≅ ΔPEN,
If you have to use SSS criterion, then you need to show
1) AR =
2) RT =
3) AT =
In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
If the following pair of the triangle is congruent? state the condition of congruency:
In ΔABC and ΔPQR, AB = PQ, AC = PR, and BC = QR.
A line segment AB is bisected at point P and through point P another line segment PQ, which is perpendicular to AB, is drawn. Show that: QA = QB.
If AP bisects angle BAC and M is any point on AP, prove that the perpendiculars drawn from M to AB and AC are equal.
In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.
If XS ⊥ QR and XT ⊥ PQ;
Prove that:
- ΔXTQ ≅ ΔXSQ.
- PX bisects angle P.
In a ΔABC, BD is the median to the side AC, BD is produced to E such that BD = DE.
Prove that: AE is parallel to BC.
