Advertisements
Advertisements
प्रश्न
In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = `(1)/(2)"AC"`.
Advertisements
उत्तर
Draw line segment DE || CB, which meets AB at point E.
Now, DE || CB and AB is the transversal,
∴ ∠AED = ∠ABC ....(corrresponding angles)
∠ABC = 90° ....(given)
⇒ ∠AED = 90°
Also, as D is the mid-point of AC and DE || CB,
DE bisects side AB,
I.e. AE = BE ....(i)
In ΔAED and ΔBED,
∠AED = ∠BED ....(Each 90°)
AE = BE ....[From (i)]
DE = DE ....(Common)
∴ ΔAED ≅ ΔBEd ....(By SAS Test)
⇒ AD = BD ....(C.P.C.T.C)
⇒ BD = AC
⇒ BD = `(1)/(2)"AC"`.
APPEARS IN
संबंधित प्रश्न
ABC is a triang D is a point on AB such that AD = `1/4` AB and E is a point on AC such that AE = `1/4` AC. Prove that DE = `1/4` BC.
BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If
L is the mid-point of BC, prove that LM = LN.
In the given figure, ΔABC is an equilateral traingle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔFED is an equilateral traingle.
L and M are the mid-point of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.
A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that
CQ = `[1]/[4]`AC. PQ produced meets BC at R.
Prove that
(i)R is the midpoint of BC
(ii) PR = `[1]/[2]` DB
In Δ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.
In triangle ABC ; D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F.
Prove that BDEF is a parallelogram. If AB = 16 cm, AC = 12 cm and BC = 18 cm,
find the perimeter of the parallelogram BDEF.
Side AC of a ABC is produced to point E so that CE = `(1)/(2)"AC"`. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meets AC at point P and EF at point R respectively. Prove that: 4CR = AB.
In ΔABC, D, E and F are the midpoints of AB, BC and AC.
If AE and DF intersect at G, and M and N are the midpoints of GB and GC respectively, prove that DMNF is a parallelogram.
In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: Q A and P are collinear.
