Advertisements
Advertisements
Question
In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that: BD = CD
Advertisements
Solution
In ΔADB and ΔADC,
∠ADB = ∠ADC ...(Since AD is perpendicular to BC)
AB = AC ...(given)
AD = AD ...(common side)
∴ ΔADB ≅ ΔADC ...(RHS congruence criterion)
⇒ BD = CD ...(cpct)
APPEARS IN
RELATED QUESTIONS
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
- ΔAMC ≅ ΔBMD
- ∠DBC is a right angle.
- ΔDBC ≅ ΔACB
- CM = `1/2` AB
You want to show that ΔART ≅ ΔPEN,
If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have
1) RT = and
2) PN =
ABCD is a square, X and Yare points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
In the given figure, prove that:
CD + DA + AB + BC > 2AC
In two congruent triangles ABC and DEF, if AB = DE and BC = EF. Name the pairs of equal angles.
In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles.
Prove that: XA = YC.
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.
ABC is an isosceles triangle with AB = AC and BD and CE are its two medians. Show that BD = CE.
ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
